Estimation Of Materials Required For LT Lines Or Low Tension Line

Hello guys, welcome back to our blog. Here in this article, we will discuss the estimation of materials required for LT lines or low tension lines, problems on LT lines or low tension lines and we will explain each and every component.

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Estimation Of Materials Required For LT Lines

01. No of spans
= Length of line ÷ span

What is the span?

Span means horizontal distance between two poles is called span. A number of spans mean for example in a 1000 KM line how many spans are there?

02. No of supports

= No of spans + 1 ( for taping pole )

Supports mean nothing but poles that give support to the conductors. A number of support means in the 1000 KM line how many supports are there.

03. R.C.C poles are used at starting, deviation points, and anchoring points

R.C.C full form is reinforced cement concrete, this pole is constructed using concrete and with some rods. This pole is very strong compared to the other pole. R.C.C pole cannot break easily, it can hold heavy weight.

04. P.C.C poles are used at intermediate poles only.

P.C.C pole’s full form is plain cement concrete, this pole is weak in tension and it is constructed only by using cement concrete.

05. No of cross-arms

No of 4-pin cross-arm = No of support + 1 ( tapping pole )

cross-arms which gives support to the insulators and conductors. cross arms are classified into many types; V shape cross arm, horizontal cross arm, U-shaped cross arm, etc.

06. No of pin insulators

= 4 [Number of supports + no of deviations – (one dead-end + anchoring points)]

Pin insulators that do not allow an electric current to flow through them, pin insulators are used if the voltage is below 33 KV. 

07. No of strain insulators

= [Two (one for each dead end) + 2(number of anchoring points )]
8. Length of the conductors required :

(a). For phases, the ACSR No.2 conductor has required = 3 × lengths of line(b). For neutral, ACSR No.4 conductor has required = 1 × length of a line
The full form of ACSR conductor is “Aluminium Conductor Steel Reinforced”. 

09. The number of guy sets

= 2(one for each dead end) + (Anchoring point)2 + deviation points.

10. Whenever guarding is to be provided two numbers of special cross arms with two numbers of guy sets and 15 kg 8 SWG G.I wire are to be used.
A guarding is provided for the safety of life, installation, and communication circuits.

The guarding for 11 KV lines is provided at road crossings, canal crossings, railway crossings, crossing over lt lines, or communication lines.

This was about materials required for LT lines.

Problem On Estimation Of Materials Required For LT Lines

Q: A three-phase 4 wire 415/230 V LT line has to be extended for a length of 1 KM, for electrification of a factory. Assuming an average span of 60 meters. Prepare the list of materials required for the same.

Solution :

1. Number of spans 

= Length of line / span

= (1 × 1000) / 60 

= 16.66 say 17 No’s

( span means the distance between two poles)  

The number of spans in the one-kilometer line is 17 No.

2. Total number of supports 

= Number of spans + 1 

= 17 + 1 = 18 No’s

The total number of supports required for a one-kilometer line is 18 No’s.

3. Number of R.C.C poles 

= one for tapping + one for dead-end + ( at 0.5 KM anchoring ) = 1 + 1 + 1 = 3 No’s

The total number of reinforced cement concrete poles required is 3 No’s.

4. Number of P.C.C poles 

= Number of supports – R.C.C poles

= 18 – 3 = 15 No’s

The number of plain cement concrete poles required is 15 No’s.

5. Number of 4-pin cross arms = 18 No’s

How many poles are there that many cross arms are required.

6. Number of 1.1 KV class guy sets

= 1(starting pole) + 1(terminating pole) + 2(at 0.5 KM anchoring pole) = 4 No’s

The number of guy set required is 4 No’s.

7. Number of pin insulators

= (No of intermediate poles + Anchoring ) × 4

= (15 + 1) × 4

= 16 × 4 = 64 No’s

The number of pin insulators required is 64 No’s

8. Number of strain insulators

= 4 * [starting pole + end pole + 2(Anchoring pole)] = 4(2 + 2) = 16 No’s

9. Length of conductors

(a). A.C.S.R No.2 weasel conductors for 3-phase conductors with 3% sag = 3 × 1 KM × 1.03 = 3.09 KM

(b). A.C.S.R No.4 squirrel conductor for neutral with 3% sag  = 1 × 1 KM × 1.03 = 1.03 KM

10. Miscellaneous materials such as nuts, bolts, barbed wires, etc.

List Of Materials Required For LT Lines

SI.No	List Of Materials	Unit	Quantity
01	R.C.C poles 7.5 meters long, 140 kg working load	No’s	03
02	P.C.C poles 7.5 meters long, 115 kg working load	No’s	15
03	4-pin cross arms with pole clamp nut, bolt, and washers	Set	18
04	1.1 KV class guy set complete	Set	04
05	1.1 KV pin insulators with pin and nuts complete set	Set	64
06	No 8 strain insulator of porcelain	No’s	16
07	No 2 A.C.S.R squirrel conductor with 3% sag for all line conductor	KM	3.09
08	No A.C.S.R squirrel conductor with 3% sag for the neutral conductor	KM	1.03
09	Miscellaneous materials such as nuts, bolts, washers, barbed wire, etc	Lump sum	45

This was about “Estimation Of Materials Required For LT Lines“. I hope this article may help you all a lot. Thank you for reading.

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